CURRENT BYU 23

Let the resistance of each series resistor be $R$ and the resistance of each voltmeter be $S$. We define the ratio $k = R/S$.
The circuit is a ladder where node voltage $V_i$ represents the reading of the $i$-th voltmeter.

Analyze Node C (Voltmeter C): Current through voltmeter C is $I_C = V_C/S$. This entire current flows from B to C through resistor $R$. $$ V_B – V_C = I_C R = \frac{V_C}{S} R = V_C k $$ $$ V_B = V_C (1 + k) \quad \dots(1) $$

Analyze Node B: Current arriving from A ($I_{AB}$) splits into current down voltmeter B ($V_B/S$) and current to C ($I_C = V_C/S$). $$ I_{AB} = \frac{V_B}{S} + \frac{V_C}{S} $$ Voltage drop across R (between A and B): $$ V_A – V_B = I_{AB} R = \left( \frac{V_B + V_C}{S} \right) R = k(V_B + V_C) $$ $$ V_A = V_B + k(V_B + V_C) = V_B(1+k) + k V_C \quad \dots(2) $$

Substitute (1) into (2): $$ V_A = [V_C(1+k)](1+k) + k V_C $$ $$ V_A = V_C (1 + 2k + k^2 + k) = V_C (k^2 + 3k + 1) $$ Given $V_A = 22, V_C = 8$: $$ 22 = 8 (k^2 + 3k + 1) $$ $$ 2.75 = k^2 + 3k + 1 $$ $$ k^2 + 3k – 1.75 = 0 $$ Solving for $k$ (must be positive): $$ k = \frac{-3 + \sqrt{9 – 4(1)(-1.75)}}{2} = \frac{-3 + \sqrt{9 + 7}}{2} = \frac{-3 + 4}{2} = 0.5 $$ Now calculate $V_B$ using eq (1): $$ V_B = 8 (1 + 0.5) = 12 \text{ V} $$